$\left(y^6\right)^3$
$6+\left\{4-\left[17-16\right]+3\right\}-5$
$\lim_{x\to2}\left(\left(x^2+4x-3\right)\right)$
$\frac{x^{2}-0.09}{x-0.3}$
$\left(4xy-3x\right)\left(4xy+3x\right)$
$\frac{4x^6-y^6}{2x^3-y^3}$
$\frac{54x^4}{3x^2}$
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